Find ΔH for: BaCO3 (s) → BaO (s) + CO2 (g) given 2 Ba (s) + O2 (g) → 2 BaO (s) ΔH = - 1107.0 kJ Ba (s) + CO2 (g) + ½ O2 (g) → BaCO3 (s) ΔH = - 822.5 kJ

The answer to your question is ΔH = 269 kJ

Explanation:

Change the sense of the third reaction

BaCO₃ (s) ⇒ Ba (s) + CO₂ (g) + 1/2O₂ (g) ΔH = 822.5 kJ

Divide the second reaction by 2

Ba (s) + 1/2O₂ (g) ⇒ BaO (s) ΔH = - 553.5 KJ

Add both reactions and cancel reactants and products

BaCO₃ (s) ⇒ BaO (s) + CO₂ (g) ΔH = 269 (822.5 - 553.5)